Question 294587 in Json Question posted in
Our archive of expertly curated questions and answers provides insights and solutions to common problems related to this popular data interchange format. From parsing and manipulating JSON data to integrating it with various programming languages and web services, our archive has got you covered. Start exploring today and take your JSON skills to the next level

Extract value from JSON array using Snowflake sql

Snowflake table has a varchar column with following JSON array as value.

[{"$":6394.2,"@":"decisionMetricValue","@id":"Month 1"},{"$":6672.25,"@":"decisionMetricValue","@id":"Month 2"},{"$":5643.05,"@":"decisionMetricValue","@id":"Month 3"},{"$":5564.79,"@":"decisionMetricValue","@id":"Month 4"},{"$":6966.75,"@":"decisionMetricValue","@id":"Month 5"},{"$":7605.15,"@":"decisionMetricValue","@id":"Month 6"}]

Using snowflake sql, I would like to extract data in following format.

Month1,Month2,Month3,Month4,Month5,Month6
6394.2,6672.25,5643.05,5564.79,6966.75,7605.15

I was trying to implement solution https://stackoverflow.com/a/77319223/20535715 , but my json is organized differently.

Thanks.

Tags:

2

Answers


  1. Chosen as BEST ANSWER
    0

    Had to use use index position differently, it works with a query as blow,

     WITH sample_data AS (
    SELECT
      PARSE_JSON('[
        {"$": 63694.1, "@": "t1"},
        {"$": 63694.2, "@": "t2"},
        {"$": 63694.3, "@": "t3"}
      ]') AS json_data
    )
    SELECT
     json_data[0]."$",
     json_data[1]."$",
     json_data[2]."$"
    FROM sample_data;

  2. 0

    You can try this one. I write it assuming that you may have multiple rows:

    with mytable as (
select ('[{"$":6394.2,"@":"decisionMetricValue","@id":"Month 1"},{"$":6672.25,"@":"decisionMetricValue","@id":"Month 2"},{"$":5643.05,"@":"decisionMetricValue","@id":"Month 3"},{"$":5564.79,"@":"decisionMetricValue","@id":"Month 4"},{"$":6966.75,"@":"decisionMetricValue","@id":"Month 5"},{"$":7605.15,"@":"decisionMetricValue","@id":"Month 6"}]' ) x
union all
select ('[{"$":6394.3,"@":"decisionMetricValue","@id":"Month 1"},{"$":6672.25,"@":"decisionMetricValue","@id":"Month 2"},{"$":5643.05,"@":"decisionMetricValue","@id":"Month 3"},{"$":5564.79,"@":"decisionMetricValue","@id":"Month 4"},{"$":6966.75,"@":"decisionMetricValue","@id":"Month 5"},{"$":7605.15,"@":"decisionMetricValue","@id":"Month 6"}]' ) x
),
flattened as (
select 
v.seq row_num,
get( v.value, '@id' )::VARCHAR colname,
get( v.value, '$' ) monvalue
from mytable,
lateral flatten (  PARSE_JSON(x) )  v
)
select * from flattened
pivot (sum(monvalue) for colname In ('Month 1','Month 2','Month 3','Month 4','Month 5'))
order by row_num
;

    This is the result:

    +---------+-----------+-----------+-----------+-----------+-----------+
| ROW_NUM | 'Month 1' | 'Month 2' | 'Month 3' | 'Month 4' | 'Month 5' |
+---------+-----------+-----------+-----------+-----------+-----------+
|       1 |    6394.2 |   6672.25 |   5643.05 |   5564.79 |   6966.75 |
|       2 |    6394.3 |   6672.25 |   5643.05 |   5564.79 |   6966.75 |
+---------+-----------+-----------+-----------+-----------+-----------+
    Login or Signup to reply.
Please signup or login to give your own answer.

Back To Top
Search